可能的重复:
MySQL 经纬度表设置
我知道这个问题可能已经被问过很多次了,我已经研究了很多,我需要特定事情的帮助.
I know this question has probably been asked many times, I've researched a lot and I need help with specific thing.
假设我有一个表单,用户输入经度和纬度,并且我有一个包含经度和纬度表的数据库,我将如何在该表中搜索半径 15 英里内的一个或多个点?
Lets say I have a form and user enters longitude and latitude, and I have a database which has table containing longitudes and latitudes, how would I search for a point or points in that table that are within 15 miles of radius?
您可以使用计算两点之间距离的公式.例如:
You can use the formula that calculates distances between two points. For example:
function get_distance($latitude1, $longitude1, $latitude2, $longitude2, $unit = 'Mi') {
$theta = $longitude1 - $longitude2;
$distance = (sin(deg2rad($latitude1)) * sin(deg2rad($latitude2))) +
(cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) *
cos(deg2rad($theta)));
$distance = acos($distance);
$distance = rad2deg($distance);
$distance = $distance * 60 * 1.1515;
switch($unit) {
case 'Mi':
break;
case 'Km' :
$distance = $distance * 1.609344;
}
return (round($distance,2));
}
您还可以执行以下操作:
You can also do something like:
$query = "SELECT *,(((acos(sin((".$latitude."*pi()/180)) *
sin((`Latitude`*pi()/180))+cos((".$latitude."*pi()/180)) *
cos((`Latitude`*pi()/180)) * cos(((".$longitude."- `Longitude`)*
pi()/180))))*180/pi())*60*1.1515
) as distance
FROM `MyTable`
HAVING distance >= ".$distance.";
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